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I am just going through the 6th chapter of Thorne and Blandford's Modern Classical Physics and got stuck on a formula (circled and red) which the authors use to calculate the center of a general wave packet.

6.2.2 Wave packets

Waves in the real world are not precisely monochromatic and planar. Instead, they occupy wave packets that are somewhat localized in space and time. Such wavepackets can be constructed as superpositions of plane waves: $$ \psi(\mathbf{x}, t)=\int \mathrm{A}(\mathbf{k}) e^{i \alpha(\mathbf{k})} e^{i(\mathbf{k} \cdot \mathbf{x}-\omega t)} d^{3} k \tag{6.7}$$ where $ \mathrm{A} $ is the modulus and $ \alpha $ the phase of the complex amplitude $ A $, and the integration element is $ d^{3} k \equiv d k_{x} d k_{y} d k_{z} $ in terms of components of $ \mathbf{k} $ on Cartesian axes $ x, y, z $. Suppose, as is often the case, that $ \mathrm{A}(\mathbf{k}) $ is sharply concentrated around some specific wave vector $ \mathbf{k}_{o} $. Then in the integral, the contributions from adjacent $\mathbf k$'s will tend to cancel each other except in that region of space and time where the oscillatory phase factor changes little with changing $ \mathbf{k} $. This is the spacetime region in which the wave packet is concentrated, and its center is where $\color{red}{\boxed{ \nabla_{\mathbf{k}}(\text{phasefactor})=0 }}$: $$ \left(\frac{\partial \alpha}{\partial k_{j}}+\frac{\partial}{\partial k_{j}}(\mathbf{k} \cdot \mathbf{x}-\omega t)\right)_{\mathbf{k}=\mathbf{k}_{o}}=0 \tag{6.8}$$ Evaluating the derivative with the aid of the wave's dispersion relation, we obtain for the location of the wave packet's center $$ \mathbf{x}_{j}-\left(\frac{\partial \Omega}{\partial k_{j}}\right)_{\mathbf{k}=\mathbf{k}_{o}} t=-\left(\frac{\partial \alpha}{\partial k_{j}}\right)_{\mathbf{k}=\mathbf{k}_{o}}=\text { const } \tag{6.9}$$

I can't find a derivation of this formula nor can I understand the physical intuition behind it. The formula states that the phase factor should not change when changing the wave vector but why is this the center of the wave packet?

They use it to derive the formula for the group velocity $$\mathbf{v}_\text{g}=\nabla_{ \mathbf{k} } \Omega$$

For further information the chapter is available online here.

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The entire phase factor is a function of $\mathbf{k}$. The center of the wave packet is where the individual waves interfere the most constructively. In order to maximize this constructive interference, the phase factors should be as close to each other as possible, and that occurs precisely at a stationary point, where the derivative of the phase factor with respect to $\mathbf{k}$ is zero. As we know from elementary calculus, there is no first-order change in a function at its stationary point.

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  • $\begingroup$ That makes sense! Thanks! $\endgroup$ Aug 1, 2021 at 13:33

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